cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A119242 Least number k such that there are exactly n powerful numbers between k^2 and (k+1)^2.

Original entry on oeis.org

1, 2, 5, 31, 234, 1822, 3611, 17329, 1511067, 524827, 180469424, 472532614, 78102676912
Offset: 0

Views

Author

T. D. Noe, May 09 2006

Keywords

Comments

Pettigrew gives a(1)-a(6) in table 14. He conjectures that k exists for every n. Surprisingly, a(8) is greater than 10^6, but a(9)=524827. The Mathematica program creates all powerful numbers <= nMax by computing all products of the form x^2 y^3.
a(10) is greater than 10^8. - Giovanni Resta, May 11 2006
a(n) > 10^11 for n >= 13. - Donovan Johnson, Sep 03 2013
Shiu (1980) proved that infinitely many values of k exist for every n. Therefore this sequence is infinite. - Amiram Eldar, Jul 10 2020

Examples

			a(3) = 31 because 968, 972 and 1000 are between 961 and 1024.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    nMax=10^12; lst={}; Do[lst=Join[lst, i^3 Range[Sqrt[nMax/i^3]]^2], {i,nMax^(1/3)}]; lst=Union[lst]; n=0; k=1; Do[n0=k; While[lst[[k]]
    				

Extensions

a(8) and the previously known a(9) from Giovanni Resta, May 11 2006
a(10)-a(11) from Donovan Johnson, Dec 07 2008
a(12) from Donovan Johnson, Sep 01 2013

A338326 The number of biquadratefree powerful numbers (A338325) between the consecutive squares n^2 and (n+1)^2.

Original entry on oeis.org

0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 2, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 3, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Amiram Eldar, Oct 22 2020

Keywords

Comments

Dehkordi (1998) proved that for each k>=0 the sequence of numbers m such that a(m) = k has a positive asymptotic density.

Examples

			a(2) = 1 since there is one biquadratefree powerful number, 8 = 2^3, between 2^2 = 4 and 3^2 = 9.
		

Crossrefs

Programs

  • Mathematica
    bqfpowQ[n_] := AllTrue[FactorInteger[n][[;; , 2]], MemberQ[{2, 3 }, #] &]; a[n_] := Count[Range[n^2 + 1, (n + 1)^2 - 1], _?bqfpowQ]; Array[a, 100]

A336175 Numbers k such that there are no powerful numbers between k^2 and (k+1)^2.

Original entry on oeis.org

1, 3, 4, 6, 7, 9, 12, 13, 17, 21, 23, 24, 26, 27, 30, 32, 34, 35, 38, 40, 43, 47, 49, 54, 60, 61, 64, 66, 68, 69, 71, 75, 80, 81, 85, 86, 91, 95, 97, 99, 105, 106, 108, 112, 120, 123, 125, 128, 131, 133, 136, 137, 139, 142, 143, 151, 153, 154, 159, 160, 162, 163
Offset: 1

Views

Author

Amiram Eldar, Jul 10 2020

Keywords

Comments

Positions of 0's in A119241.
Shiu (1980) proved that this sequence has an asymptotic density 0.2759... A more accurate calculation using his formula gives 0.275965511407718981...

Examples

			1 is a term since the two numbers between 1^2 = 1 and (1+1)^2 = 4, 2 and 3, are not powerful.
2 is not a term since there is a powerful number, 8 = 2^3, between 2^2 = 4 and (2+1)^2 = 9.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[150], ! AnyTrue[Range[#^2 + 1, (# + 1)^2 - 1], powQ] &]
  • PARI
    is(n)=forfactored(k=n^2+1,n^2+2*n, if(ispowerful(k), return(0))); 1 \\ Charles R Greathouse IV, Oct 31 2022
    
  • Python
    from functools import lru_cache
    from math import isqrt
    from sympy import mobius, integer_nthroot
    def A336175(n):
        def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        @lru_cache(maxsize=None)
        def g(x):
            c, l = 0, 0
            j = isqrt(x)
            while j>1:
                k2 = integer_nthroot(x//j**2,3)[0]+1
                w = squarefreepi(k2-1)
                c += j*(w-l)
                l, j = w, isqrt(x//k2**3)
            c += squarefreepi(integer_nthroot(x,3)[0])-l
            return c
        def f(x):
            c, a = n+x, 1
            for k in range(1,x+1):
                b = g((k+1)**2)
                if b == a+1:
                    c -= 1
                a = b
            return c
        return bisection(f,n,n) # Chai Wah Wu, Sep 14 2024

Formula

a(n) ~ k*n, where k = 3.623641211175... is the inverse of the density, see Shiu link. - Charles R Greathouse IV, Oct 31 2022

A337736 The number of cubefull numbers (A036966) between the consecutive cubes n^3 and (n+1)^3.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 0, 4, 2, 1, 3, 0, 3, 1, 2, 1, 3, 2, 0, 2, 5, 1, 3, 1, 1, 3, 3, 2, 1, 3, 1, 2, 2, 2, 1, 2, 2, 3, 6, 1, 1, 1, 4, 1, 1, 3, 3, 1, 3, 4, 1, 2, 3, 1, 2, 3, 2, 3, 2, 3, 3, 2, 1, 4, 2, 1, 1, 0, 7, 1, 1, 4, 3, 2, 2, 2, 3, 3, 2, 0, 4, 2, 4
Offset: 1

Views

Author

Amiram Eldar, Sep 17 2020

Keywords

Comments

For each k >= 0 the sequence of solutions to a(x) = k has a positive asymptotic density (Shiu, 1991).

Examples

			a(2) = 1 since there is one cubefull number, 16 = 2^4, between 2^3 = 8 and 3^3 = 27.
		

Crossrefs

Programs

  • Mathematica
    cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; a[n_] := Count[Range[n^3 + 1, (n + 1)^3 - 1], _?cubQ]; Array[a, 100]
  • Python
    from math import gcd
    from sympy import integer_nthroot, factorint
    def A337736(n):
        def f(x):
            c = 0
            for w in range(1,integer_nthroot(x,5)[0]+1):
                if all(d<=1 for d in factorint(w).values()):
                    for y in range(1,integer_nthroot(z:=x//w**5,4)[0]+1):
                        if gcd(w,y)==1 and all(d<=1 for d in factorint(y).values()):
                            c += integer_nthroot(z//y**4,3)[0]
            return c
        return f((n+1)**3-1)-f(n**3) # Chai Wah Wu, Sep 10 2024

Formula

Asymptotic mean: lim_{m->oo} (1/m) Sum_{k=1..m} a(k) = A362974 - 1 = 3.659266... . - Amiram Eldar, May 11 2023

A336176 Numbers k such that there is a single powerful number between k^2 and (k+1)^2.

Original entry on oeis.org

2, 8, 10, 15, 16, 18, 19, 20, 28, 29, 37, 39, 41, 42, 45, 48, 50, 51, 52, 53, 56, 57, 59, 63, 65, 74, 76, 77, 78, 79, 83, 84, 87, 89, 90, 92, 94, 100, 101, 102, 107, 113, 114, 115, 116, 117, 118, 119, 121, 122, 126, 127, 130, 134, 138, 141, 144, 146, 147, 148
Offset: 1

Views

Author

Amiram Eldar, Jul 10 2020

Keywords

Comments

Positions of 1's in A119241.
Shiu (1980) proved that this sequence has an asymptotic density 0.3955... A more accurate calculation using his formula gives 0.3955652153962362...
1 is the most common value of A119241.

Examples

			2 is a term since there is a single powerful number, 8 = 2^3, between 2^2 = 4 and (2+1)^2 = 9.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[150], Count[Range[#^2 + 1, (# + 1)^2 - 1], _?powQ] == 1 &]

A336177 Numbers k such that there are exactly two powerful numbers between k^2 and (k+1)^2.

Original entry on oeis.org

5, 11, 14, 22, 25, 33, 44, 46, 55, 58, 62, 70, 72, 73, 82, 88, 96, 98, 103, 104, 109, 110, 111, 124, 129, 135, 155, 156, 158, 164, 172, 176, 178, 181, 187, 197, 203, 206, 207, 209, 212, 218, 240, 243, 248, 249, 254, 257, 259, 268, 277, 279, 281, 285, 288, 291
Offset: 1

Views

Author

Amiram Eldar, Jul 10 2020

Keywords

Comments

Positions of 2's in A119241.
Shiu (1980) proved that this sequence has an asymptotic density 0.2312... A more accurate calculation using his formula gives 0.231299167354828...

Examples

			5 is a term since there are exactly two powerful numbers, 27 = 3^3 and 32 = 2^5 between 5^2 = 25 and (5+1)^2 = 36.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[300], Count[Range[#^2 + 1, (# + 1)^2 - 1], _?powQ] == 2 &]

A336178 Numbers k such that there are exactly three powerful numbers between k^2 and (k+1)^2.

Original entry on oeis.org

31, 36, 67, 93, 132, 140, 145, 161, 166, 189, 192, 220, 223, 265, 280, 290, 296, 311, 316, 322, 364, 384, 407, 468, 537, 576, 592, 602, 623, 639, 644, 656, 659, 661, 670, 690, 722, 769, 771, 793, 828, 883, 888, 890, 896, 950, 961, 981, 984, 987, 992, 995, 1018
Offset: 1

Views

Author

Amiram Eldar, Jul 10 2020

Keywords

Comments

Positions of 3's in A119241.
Shiu (1980) proved that this sequence has an asymptotic density = 0.0770... A more accurate calculation using his formula gives 0.0770742722233...

Examples

			31 is a term since there are exactly three powerful numbers, 968 = 2^3 * 11^2, 972 = 2^2 * 3^5 and 1000 = 2^3 * 5^3 between 31^2 = 961 and (31+1)^2 = 1024.
		

References

  • József Sándor, Dragoslav S. Mitrinovic and Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, chapter VI, p. 226.

Crossrefs

Programs

  • Mathematica
    powQ[n_] := (n == 1) || Min @@ FactorInteger[n][[;; , 2]] > 1; Select[Range[1000], Count[Range[#^2 + 1, (# + 1)^2 - 1], _?powQ] == 3 &]
  • Python
    from functools import lru_cache
    from math import isqrt
    from sympy import mobius, integer_nthroot
    def A336178(n):
        def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1)))
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        @lru_cache(maxsize=None)
        def g(x):
            c, l = 0, 0
            j = isqrt(x)
            while j>1:
                k2 = integer_nthroot(x//j**2,3)[0]+1
                w = squarefreepi(k2-1)
                c += j*(w-l)
                l, j = w, isqrt(x//k2**3)
            c += squarefreepi(integer_nthroot(x,3)[0])-l
            return c
        def f(x):
            c, a = n+x, 1
            for k in range(1,x+1):
                b = g((k+1)**2)
                if b == a+4:
                    c -= 1
                a = b
            return c
        return bisection(f,n,n) # Chai Wah Wu, Sep 14 2024

A361936 Indices of the squares in the sequence of powerful numbers (A001694).

Original entry on oeis.org

1, 2, 4, 5, 6, 9, 10, 11, 13, 14, 16, 19, 20, 21, 24, 26, 28, 29, 31, 33, 35, 36, 39, 40, 41, 44, 45, 46, 48, 50, 51, 55, 56, 59, 60, 61, 65, 67, 68, 70, 71, 73, 75, 76, 79, 81, 84, 85, 87, 88, 90, 92, 94, 96, 97, 100, 102, 104, 107, 109, 110, 111, 114, 116, 117, 119, 120
Offset: 1

Views

Author

Amiram Eldar, Mar 31 2023

Keywords

Comments

Equivalently, the number of powerful numbers that do not exceed n^2.
The asymptotic density of this sequence is zeta(3)/zeta(3/2) = 1/A090699 = 0.460139... .
If k is a term of A336175 then a(k) and a(k+1) are consecutive integers, i.e., a(k+1) = a(k) + 1.

Crossrefs

Programs

  • Mathematica
    Position[Select[Range[5000], # == 1 || Min[FactorInteger[#][[;; , 2]]] > 1 &], _?(IntegerQ[Sqrt[#]] &)] // Flatten
  • PARI
    lista(kmax) = {my(c = 0); for(k = 1, kmax, if(ispowerful(k), c++); if(issquare(k), print1(c, ", "))); }
    
  • Python
    from math import isqrt
    from sympy import integer_nthroot, factorint
    def A361936(n):
        m = n**2
        return int(sum(isqrt(m//k**3) for k in range(1, integer_nthroot(m, 3)[0]+1) if all(d<=1 for d in factorint(k).values()))) # Chai Wah Wu, Sep 10 2024

Formula

a(n) = A217038(n^2).
a(n+1) - a(n) = A119241(n) + 1.
a(n) = (zeta(3/2)/zeta(3)) * n + O(n^(2/3)).
Showing 1-8 of 8 results.